Question

The derivative of \( y=\int_{0}^{\ln x}\sin(e^t)\,dt \) is :

• A. \(1\)
• B. \(\dfrac{\sin x}{x}\)
• C. \(\dfrac{\cos x}{x}\)
• D. \(1-\cos e^x\)

Hint:

For integrals with variable upper limit: \[ \frac{d}{dx}\left(\int_a^{g(x)}f(t)\,dt\right) = f(g(x))\cdot g'(x) \] Always:
• substitute the upper limit into the integrand,
• then multiply by derivative of the upper limit.

Answer 1

The Correct Option is B

Concept: This problem is based on the Fundamental Theorem of Calculus together with the Chain Rule. If \[ F(x)=\int_{a}^{g(x)} f(t)\,dt \] then: \[ \frac{d}{dx}F(x)=f(g(x))\cdot g'(x) \] Thus:
• First substitute the upper limit into the integrand,
• Then differentiate the upper limit,
• Finally multiply both results. This technique is extremely important for differentiating variable limit integrals.

Step 1: Writing the given integral carefully. We are given: \[ y=\int_{0}^{\ln x}\sin(e^t)\,dt \] Here: \[ f(t)=\sin(e^t) \] and the upper limit is: \[ g(x)=\ln x \]

Step 2: Applying the Fundamental Theorem of Calculus. Using: \[ \frac{d}{dx}\left(\int_{a}^{g(x)} f(t)\,dt\right) = f(g(x))\cdot g'(x) \] we get: \[ \frac{dy}{dx} = \sin(e^{\ln x})\cdot \frac{d}{dx}(\ln x) \]

Step 3: Simplifying \( e^{\ln x} \). Using the logarithmic-exponential identity: \[ e^{\ln x}=x \] Therefore: \[ \sin(e^{\ln x})=\sin x \] Hence: \[ \frac{dy}{dx} = \sin x \cdot \frac{1}{x} \]

Step 4: Obtaining the final derivative. Thus: \[ \boxed{ \frac{dy}{dx} = \frac{\sin x}{x} } \] Hence the correct option is: \[ \boxed{(2)\;\dfrac{\sin x}{x}} \]

Step 5: Verifying the remaining options.
• Option \(1\): \(1\) \(\rightarrow\) Incorrect because derivative depends on \(x\)
• Option \(2\): \(\dfrac{\sin x}{x}\) \(\rightarrow\) Correct
• Option \(3\): \(\dfrac{\cos x}{x}\) \(\rightarrow\) Incorrect because integrand involves \(\sin(e^t)\)
• Option \(4\): \(1-\cos e^x\) \(\rightarrow\) Incorrect form